Define a cylinder: $S = \{ (x, y, z) \in \mathbb{R}^3 \big | x^2 + y^2 \leq 3, \; -2 \leq z \leq 1 \}$ What is the triple integral of the scalar field $f(r, \theta) = r^2\cos(\theta)$ over $S$ in cylindrical coordinates? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{-2}^1 \int_0^3 \int_0^{2\pi} r^3\cos(\theta) \, d\theta \, dr \, dz$ (Choice B) B $\dfrac{1}{2} \int_{-2}^1 \int_0^3 \int_0^{2\pi} r^3\sin(2\theta) \, d\theta \, dr \, dz$ (Choice C) C $ \int_0^1 \int_0^3 \int_0^{2\pi} r^2\cos(\theta) \, d\theta \, dr \, dz$ (Choice D) D $ \int_{-2}^1 \int_0^3 \int_0^{2\pi} r^4\cos(\theta)\sin(\theta) \, d\theta \, dr \, dz$
Explanation: The only bound is $0 < \theta < 2\pi$. Here is the change of variables for cylindrical coordinates. $\begin{aligned} x &= r \cos(\theta) \\ \\ y &= r \sin(\theta) \\ \\ z &= z \end{aligned}$ We want to represent the cylinder $S$ with bounds in cylindrical coordinates. Here, the region $S$ ranges from a height of $z = -2$ to $z = 1$ and radius of $r = 0$ to $r = 3$. Theta goes from $0$ to $2\pi$ so that we integrate over the whole cylinder and not just a wedge. $ \int_{-2}^1 \int_0^3 \int_0^{2\pi} \cdots \, d\theta \, dr \, dz$ The scalar field we want to integrate over depends the distance from the origin $r$ and the angle $\theta$, so we don't need to make any substitutions with cylindrical coordinates. Our integral looks like this now: $ \int_{-2}^1 \int_0^3 \int_0^{2\pi} r^2\cos(\theta) \cdots \, d\theta \, dr \, dz$ The final step is finding the Jacobian of cylindrical coordinates, which we'll need to multiply in to get the final integral. $J(r, \theta, z) = r$ [Derivation] The integral in cylindrical coordinates: $ \int_{-2}^1 \int_0^3 \int_0^{2\pi} r^3\cos(\theta) \, d\theta \, dr \, dz$